CODEWITHSUNDEEP
operatorsocaml
Mike Samuel
Mike Samuel

Reputation: 120516

Referencing operator ::

I'm used to using ( + ) to reference the pervasive integer addition operator, but this doesn't work for ( :: ):

        OCaml version 4.01.0

# (+);;
- : int -> int -> int = <fun>
# ( :: );;
Error: Syntax error: operator expected.
# ( := );;
- : 'a ref -> 'a -> unit = <fun>

The expression grammar says

expr  ::= ...
        | expr  ::  expr
        ...
        ∣ [ expr  { ; expr }  [;] ]  
        ...
        | expr  infix-op  expr
        ...

and lexical conventions says

infix-symbol  ::=  (= ∣  < ∣  > ∣  @ ∣  ^ ∣  | ∣  & ∣  + ∣  - ∣  * ∣  / ∣  $ ∣  %) { operator-char }

which seems to exclude both :: and := as infix operators, even though ( := ) works just fine.

What is ::'s status as an operator?

Is there a convenient handle for the list prepend operator or is (fun el ls -> el::ls) the best one can do?

Upvotes: 1

Views: 152

Answers (3)

newacct
newacct

Reputation: 122439

which seems to exclude both :: and := as infix operators, even though ( := ) works just fine.

You made a jump there. You cite

expr  ::= ...
        | expr  infix-op  expr

But then you did not look at infix-op, which is defined as

infix-op    ::= infix-symbol  
         ∣   * ∣  + ∣  - ∣  -. ∣  = ∣  != ∣  < ∣  > ∣  or ∣  || ∣  & ∣  && ∣  :=  
         ∣   mod ∣  land ∣  lor ∣  lxor ∣  lsl ∣  lsr ∣  asr

So there, := is a infix operator, along with other ones like mod, etc. infix-symbol is only for custom infix operators.

Upvotes: 2

sandymatt
sandymatt

Reputation: 5612

The cons operator :: is a constructor, and it can't be applied as an infix operator.

Have a look at the pervasive module for a list of all the ones you can use like that.

Upvotes: 2

Thomash
Thomash

Reputation: 6379

:: is a base construction that can't be implemented with other constructions. := is an operator that can be implemented as let (:=) r v = r.contents <- v. But I agree that this contradicts the lexical conventions described in the manual.

For your problem of using (::), the best you can do is to give it a short name if you want to use it multiple times. let cons h t = h :: t

Upvotes: 2

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